Integral of \( \sec x \)

Derivation and Evaluation

Evaluate the integral:

\[ \int \sec x \, dx \]

Use \( \sec x = \dfrac{1}{\cos x} \) to rewrite the integral:

\[ \int \sec x \, dx = \int \dfrac{1}{\cos x} \, dx \]

Multiply both numerator and denominator of the integrand on the right by \( \cos x \):

\[ \int \sec x \, dx = \int \dfrac{\cos x}{\cos^2 x} \, dx \]

Use the trigonometric identity \( \cos^2 x = 1 - \sin^2 x \) in the denominator to write:

\[ \int \sec x \, dx = \int \dfrac{\cos x}{1 - \sin^2 x} \, dx \]

Use Integration by Substitution: let \( u = \sin x \), which gives \( du = \cos x \, dx \). The integral can be written as:

\[ \int \sec x \, dx = \int \dfrac{1}{1 - u^2} \, du \]

Use partial fraction decomposition to write the integrand as:

\[ \dfrac{1}{1 - u^2} = \dfrac{1}{2(u + 1)} - \dfrac{1}{2(u - 1)} \]

Substitute this into the integral:

\[ \int \sec x \, dx = \dfrac{1}{2} \int \dfrac{1}{u + 1} \, du - \dfrac{1}{2} \int \dfrac{1}{u - 1} \, du \]

Use the integration formula \( \displaystyle \int \dfrac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c \) to obtain:

\[ \int \sec x \, dx = \dfrac{1}{2} \ln|u + 1| - \dfrac{1}{2} \ln|u - 1| + c \]

Group the logarithmic expressions using the property \( \ln a - \ln b = \ln\left(\dfrac{a}{b}\right) \):

\[ \int \sec x \, dx = \dfrac{1}{2} \ln\left(\dfrac{|u + 1|}{|u - 1|}\right) + c \]

Substitute back \( u = \sin x \):